What is the derivative of 6/((s^3-2)^3) ?

The derivative of 6/((s^3-2)^3) is -(54s^2)/((s^3-2)^4)

What is the first derivative of 6/((s^3-2)^3) ?

The first derivative of 6/((s^3-2)^3) is -(54s^2)/((s^3-2)^4)

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derivative of

\frac{6}{( s ^{3} - 2 ) ^{3}}

- \frac{54 s ^{2}}{( s ^{3} - 2 ) ^{4}}

Solution steps

\frac{d}{d s} (\frac{6}{( s ^{3} - 2 ) ^{3}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d s} (\frac{1}{( s ^{3} - 2 ) ^{3}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 6 \frac{d}{d s} ((s^{3} - 2)^{- 3})

Apply the chain rule:

- \frac{3}{( s ^{3} - 2 ) ^{4}} \frac{d}{d s} (s^{3} - 2)

= - \frac{3}{( s ^{3} - 2 ) ^{4}} \frac{d}{d s} (s^{3} - 2)

\frac{d}{d s} (s^{3} - 2) = 3 s^{2}

= 6 (- \frac{3}{( s ^{3} - 2 ) ^{4}} \cdot 3 s^{2})

Simplify

6 (- \frac{3}{( s ^{3} - 2 ) ^{4}} \cdot 3 s^{2}) : - \frac{54 s ^{2}}{( s ^{3} - 2 ) ^{4}}

= - \frac{54 s ^{2}}{( s ^{3} - 2 ) ^{4}}

\frac{\partial}{\partial y} (x - \frac{1}{y ^{2}})

y = - x^{2} + 4, y = - \frac{x ^{2}}{2} + 2

\frac{1}{( s ^{2} + s + 1 )}

x \to 5 lim (4 x - 3)

\int \frac{ln ( ln ( x ) )}{x ln ( x )} d x

What is the derivative of 6/((s^3-2)^3) ?
The derivative of 6/((s^3-2)^3) is -(54s^2)/((s^3-2)^4)
What is the first derivative of 6/((s^3-2)^3) ?
The first derivative of 6/((s^3-2)^3) is -(54s^2)/((s^3-2)^4)